Find the sum of all the terms in the nth row of the series given below.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 .......................... ............................ (so on)
Input : n = 2 Output : 18 terms in 2nd row and their sum sum = (3 + 4 + 5 + 6) = 18 Input : n = 4 Output : 132
Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum.
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula:
Sum(n) = n * (2 * n2 + 1)
The proof for the formula is given below:
- Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as:
Sum = (n * [2*a + (n-1)*d]) / 2
- Sum of 1st n natural numbers is given as:
Sum = (n * (n + 1)) / 2
Let the number of terms from the beginning till the end of the nth row be p. Here p = 2 + 4 + 6 + .....n terms For the given AP series, a = 2, d = 2. Using the above formula for the sum of n terms of the AP series, we get, p = n * (n + 1) Similarly, let the number of terms from the beginning till the end of the (n-1)th row be q. Here q = 2 + 4 + 6 + .....n-1 terms For the given AP series, a = 2, d = 2. Using the above formula for the sum of n-1 terms of the AP series, we get, q = n * (n - 1) Now, Sum of all the terms in the nth row = sum of 1st p natural numbers - sum of 1st q natural numbers = (p * (p + 1)) / 2 - (q * (q + 1)) / 2 Substituting the values of p and q and then solving the equation, we will get, Sum of all the terms in the nth row = n * (2 * n2 + 1)
Sum of all the terms in nth row = 132
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